Question

Difficulty: MediumDivisibility, Factors, and Multiples

If mm and nn are positive integers such that 3m5n3^m \cdot 5^n is a factor of 15!15!, what is the maximum possible value of m+nm + n?

  1. A
    8
  2. 9Answer
  3. C
    10
  4. D
    11
  5. E
    12

Answer

9
To find the maximum power of a prime pp dividing n!n!, we sum n/pk\lfloor n/p^k \rfloor for all k1k \ge 1. For p=3p = 3, m=15/3+15/9=5+1=6m = \lfloor 15/3 \rfloor + \lfloor 15/9 \rfloor = 5 + 1 = 6. For p=5p = 5, n=15/5=3n = \lfloor 15/5 \rfloor = 3. Thus, the maximum value of m+nm + n is 6+3=96 + 3 = 9.

Step-by-Step Solution

1
Find the maximum exponent mm of the prime factor 3 in 15!15! using Legendre's formula.
m=153+1532=5+1=6m = \lfloor \frac{15}{3} \rfloor + \lfloor \frac{15}{3^2} \rfloor = 5 + 1 = 6
The prime factor 3 appears in multiples of 3 (3, 6, 9, 12, 15) and contributes an extra factor in 9 (323^2).
2
Find the maximum exponent nn of the prime factor 5 in 15!15! using Legendre's formula.
n=155=3n = \lfloor \frac{15}{5} \rfloor = 3
The prime factor 5 appears in multiples of 5 (5, 10, 15).
3
Sum the maximum possible integer values of mm and nn.
m+n=6+3=9m + n = 6 + 3 = 9
The maximum possible value of m+nm+n is the sum of the maximum individual prime exponents.

Key Concept

Counting Prime Factors in a Factorial (Legendre's Formula)

Alternative Method

List out the prime factorizations of all numbers from 1 to 15: 3 contributing numbers are 3, 6 (232 \cdot 3), 9 (323^2), 12 (2232^2 \cdot 3), 15 (353 \cdot 5). Total factors of 3 = 1+1+2+1+1=61 + 1 + 2 + 1 + 1 = 6. 5 contributing numbers are 5, 10 (252 \cdot 5), 15 (353 \cdot 5). Total factors of 5 = 1+1+1=31 + 1 + 1 = 3. Sum 6+3=96 + 3 = 9.
Estimated Time:1m 30s
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